RS Aggarwal 2019,2020 solution class 7 chapter 4 Rational Numbers Exercise 4F

Exercise 4F

Page-78

Question 1:

Find the multiplicative inverse of reciprocal of each of the following:

(i) 18
(ii) −16
(iii) $\frac{13}{25}$
(iv) $\frac{-17}{12}$
(v) $\frac{-6}{19}$
(vi) $\frac{-3}{-5}$
(vii) −1
(viii) 0

Answer 1:

$$\begin{gathered} \begin{array}{l}\left(\text{i}\right)\text{ Multiplicative inverse of 18 }=\frac{1}{18}\\ \end{array} \\ \begin{array}{l}\text{ii}) \text{Multiplicative inverse of}-\text{16 }=\frac{-1}{16}\\ \end{array} \\ \begin{array}{l}\text{iii})\text{ Multiplicative inverse of }\frac{13}{25}\\ \end{array}=\frac{25}{13} \\ \begin{array}{l}\text{iv})\text{ Multiplicative inverse of}\frac{-17}{12}=\frac{12}{-17}\\ \end{array} \\ \begin{array}{l}\text{v}) \text{Multiplicative inverse of}\frac{-6}{19}=\frac{19}{-6}\\ \end{array} \\ \begin{array}{l}\text{vi}) \text{Multiplicative inverse of}\frac{-3}{-5}=\frac{-5}{-3}=\frac{5}{3}\\ \end{array} \\ \begin{array}{l}\text{vii}) \text{Multiplicative inverse of}-1 =\frac{1}{-1}=-1\\ \end{array} \\ \begin{array}{l}\text{viii}) \text{Multiplicative inverse of }0 = \frac{1}{0}=\text{infinity}\\ \\ \text{Hence, it does not exist}\text{. }\end{array} \end{gathered}$$

Question 2:

Simplify:

(i) $\frac{4}{9}÷\left(\frac{-5}{12}\right)$
(ii) $-8÷\left(\frac{-5}{16}\right)$
(iii) $\frac{-12}{7}÷\left(-18\right)$
(iv) $\left(\frac{-1}{10}\right)÷\left(\frac{-8}{5}\right)$
(v) $\left(\frac{-16}{35}\right)÷\left(\frac{-15}{14}\right)$
(vi) $\left(\frac{-65}{14}\right)÷\left(\frac{13}{-7}\right)$

Answer 2:

$\begin{array}{l}\text{(i})\frac{4}{9}÷(\frac{-5}{12})\\ =\frac{4}{{\cancel{9}}_3}\times \frac{\sout{1}{\sout{2}}^4}{(-5)}\\ =\frac{4\times 4}{3\times (-5)}\\ =\frac{-16}{15}\\ \text{(ii})-8÷(\frac{-5}{16})\\ =-8\times \frac{-16}{5}\\ =\frac{128}{5}\\ \text{(iii})\frac{-12}{7}÷(-18)\\ =\frac{-\sout{1}{\sout{2}}^2}{7}\times \left(\frac{-1}{\sout{1}{\sout{8}}_3}\right)\\ =\frac{2}{21}\\ \text{(iv})\frac{-1}{10}÷(\frac{-8}{5})\\ =\frac{-1}{\sout{1}{\sout{0}}^2}\times \left(\frac{\sout{5}}{-8}\right)\\ =\frac{-1}{2}\times \frac{1}{(-8)}\\ =\frac{-1}{-16}\\ =\frac{1}{16}\\ \text{(v})\frac{-16}{35}÷(\frac{-15}{14})\\ =\frac{-16}{\sout{3}{\sout{5}}_5}\times \frac{\sout{1}{\sout{4}}^2}{(-15)}\\ =\frac{-32}{-75}\\ =\frac{32}{75}\\ \text{(vi})\left(\frac{-65}{14}\right)÷\left(\frac{13}{-7}\right)\\ =\left(\frac{-\sout{6}{\sout{5}}^5}{\sout{1}{\sout{4}}_2}\right)\times \frac{(-\sout{7})}{\sout{1}\sout{3}}\\ =\left(\frac{-5}{2}\right)\times \left(\frac{-1}{1}\right)\\ =\frac{5}{2}\end{array}$

Question 3:

Fill in the blanks:

(i) (......) ÷ $\left(\frac{-7}{5}\right)=\frac{10}{19}$
(ii) (......) ÷ $\left(-3\right)=\frac{-4}{15}$
(iii) $\frac{9}{8}÷(......)=\frac{-3}{2}$
(iv) $\left(-12\right)÷\left(......\right)=\frac{-6}{5}$

Answer 3:

$\begin{array}{l}\text{(i})(\dots ?..)÷\frac{-7}{5}=\frac{10}{19}\\ (\dots ?..)=\frac{10}{19}\times \frac{-7}{5}\\ (\dots ?..)=\frac{-14}{19}\\ \begin{array}{l}\text{(ii}) (..?\mathrm{..})÷(-3)=\frac{-4}{15}\\ (\dots ?..)=\frac{-4}{15}\times (-3)\\ (\dots ?..)=\frac{4}{5}\\ \begin{array}{l}\text{(iii)}\frac{9}{8}÷(..?..)\text{ =}\frac{-3}{2}\\ \frac{9}{8}÷(..?..)=\frac{(-3)}{2}\end{array}\end{array}\\ (..?..)\text{ =}\frac{9}{8}\times \frac{2}{(-3)}\\ (..?..)=\frac{-3}{4}\\ \\ \text{(iv}) (-12)÷(..?..)=\frac{-6}{5}\\ (..?..)=(-12)\times \left(\frac{5}{-6}\right)\\ (..?..)=10\end{array}$

Question 4:

Divide the sum of $\frac{65}{12}$ and $\frac{8}{3}$ by their difference.

Answer 4:

$\begin{array}{l}\text{Sum =}\frac{65}{12}+\frac{8}{3}=\frac{65+32}{12}=\frac{97}{12}\\ \text{Difference}=\frac{65}{12}-\frac{8}{3}=\frac{65-32}{12}=\frac{33}{12}\\ \frac{97}{12}÷\frac{33}{12}\\ =\frac{97}{\sout{1}{\sout{2}}_1}\times \frac{\sout{1}{\sout{2}}^1}{33}\\ \text{ = }\frac{97}{33}\end{array}$

Question 5:

By what number should $\frac{-44}{9}$ be divided to get $\frac{-11}{3}$?

Answer 5:

$\begin{array}{l}\text{Let the required number be} x.\\ \begin{array}{l}\frac{-44}{9}÷x = \frac{-11}{3}\\ =>x=\frac{-\sout{4}{\sout{4}}^4}{{\sout{9}}_3}\times \frac{\sout{3}}{-\sout{11}}\\ =>x=\frac{4}{3}\end{array}\\ \end{array}$    
    

Question 6:

By what number should $\frac{-8}{15}$ be multiplied to get 24?

Answer 6:

$\begin{array}{l}\\ \text{Let the required number be }\mathit{\text{x.}}\\ x \times \left(\frac{-8}{15}\right)=24\\ \begin{array}{l}x=\text{24}÷\frac{-8}{15}\\ =\sout{2}{\sout{4}}^3\times \left(\frac{15}{-{\sout{8}}_1}\right)\\ =\frac{45\times (-1)}{-1\times (-1)}\\ =-45\end{array}\end{array}$

Question 7:

The product of two rational numbers is 10. If one of the numbers is −8, find the other.

Answer 7:

$\begin{array}{l}\text{Let the other number} \mathrm{be} x.\\ x \times -8=10\\ \begin{array}{l}=>x=\text{1}0÷(-\text{8})\\ =10\times \frac{1}{-8}\\ =\sout{1}{\sout{0}}^5\times \frac{1}{-{\sout{8}}_4}\end{array}\\ \\ \therefore \mathrm{Other} \mathrm{number} =\frac{-5}{4}\end{array}$

Question 8:

The product of two rational numbers is −9. If one of the numbers is −12, find the other.

Answer 8:

$\begin{array}{l}\text{Let the other number be x. }\\ x \times (-12) =-9\\ \begin{array}{l}=>x=-\text{9}÷(-\text{12}) \\ =-\text{9}\times \left(\frac{1}{-12}\right)\\ =-\text{9}\times \left(\frac{-1}{12}\right)\\ =\frac{{\sout{9}}^3}{\sout{1}{\sout{2}}_4}\end{array}\\ \mathrm{Hence}, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{3}{4}.\end{array}$

Question 9:

The product of two rational numbers is $\frac{-16}{9}$. If one of the numbers is $\frac{-4}{3}$, find the other.

Answer 9:

$\begin{array}{l}\text{Let the other number be x.}\\ x \times \left(\frac{-4}{3}\right)=\frac{-16}{9}\\ \begin{array}{l}\begin{array}{l}=> x= \frac{-16}{9}÷\left(\frac{-4}{3}\right)\\ =\frac{-\sout{1}{\sout{6}}^4}{{\cancel{9}}_3}\times \left(\frac{{\cancel{3}}^1}{-{\cancel{4}}_1}\right)\end{array}\\ =\frac{4}{3}\end{array}\\ \mathrm{Hence}, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{4}{3}. \end{array}$

Question 10:

By what rational number should $\frac{-8}{39}$ be multiplied to obtain $\frac{5}{26}$?

Answer 10:

$\begin{array}{l}\mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{be} x.\\ \begin{array}{l}x \times \left(\frac{-8}{39}\right)=\frac{5}{26}\\ \begin{array}{l} => x=\frac{5}{26}÷\left(\frac{-8}{39}\right)\\ \text{ = }\frac{5}{\sout{2}{\sout{6}}_2}\times \left(\frac{\sout{3}{\sout{9}}^3}{-8}\right)\\ \text{ =}\frac{15\times -1}{-16\times -1}\text{=}\frac{-15}{16}\end{array}\\ \mathrm{Hence}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{is} \frac{-15}{16}.\end{array}\\ \end{array}$

Question 11:

If 24 pairs of trousers of equal size can be prepared with 54 m of cloth, what length of cloth is required for each pair of trousers?

Answer 11:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{cloth} \mathrm{required} \mathrm{to} \mathrm{prepare} 24 \mathrm{trousers} = 54 \mathrm{m} \\ \begin{array}{l}\mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{cloth} \mathrm{required} \mathrm{for} \mathrm{each} \mathrm{pair} \mathrm{of} \mathrm{trouser}s= 54 ÷ 24\\ \begin{array}{l} =\frac{\sout{5}{\sout{\sout{4}}}^9}{\sout{2}{\sout{4}}_4}\\ \text{ =}\frac{9}{4}\text{=2}\frac{1}{4}\text{m}\end{array}\end{array} \\ \mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}, 2\frac{1}{4 } \mathrm{m} \mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h} \mathrm{o}\mathrm{f} \mathrm{c}\mathrm{l}\mathrm{o}\mathrm{t}\mathrm{h} \mathrm{is} \mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{d} \mathrm{f}\mathrm{o}\mathrm{r} \mathrm{e}\mathrm{a}\mathrm{c}\mathrm{h} \mathrm{p}\mathrm{a}\mathrm{i}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{t}\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{e}\mathrm{rs}. \end{gathered}$$

Question 12:

How many pieces, each of length $3\frac{3}{4}$ m, can be cut from a rope of length 30 m?

Answer 12:

$\begin{array}{l}\mathrm{Length} \mathrm{of} \mathrm{a} \mathrm{rope} = 30 \mathrm{m}\\ \mathrm{Number} \mathrm{of} \mathrm{pieces}=30 ÷ 3\frac{3}{4}=30 ÷ \frac{15}{4}\\ \begin{array}{l} {=}^2\sout{3}\sout{0} \times \frac{4}{\sout{1}\sout{5}}\\ =8\end{array}\\ \\ \text{ Hence, the number of pieces would be 8. }\end{array}$

Question 13:

The cost of $2\frac{1}{2}$ metres of cloth is Rs $78\frac{3}{4}$. Find the cost of cloth per metre.

Answer 13:

$\begin{array}{l}\mathrm{Cost} \mathrm{of} 2\frac{1}{2} \mathrm{m} \mathrm{cloth} = \mathrm{Rs} 78\frac{3}{4}\\ \begin{array}{l}\text{Cost of cloth per meter }=78\frac{3}{4} ÷ 2\frac{1}{2}\\ =\frac{315}{4} ÷ \frac{5}{2}\\ =\frac{\sout{3}\sout{1}{\sout{5}}^{63}}{{\cancel{4}}_2} \times \frac{{\cancel{2}}^1}{{\sout{5}}_1}\\ =\text{Rs }\frac{63}{2}=Rs 31\frac{1}{2}\end{array}\\ \therefore \mathrm{Cost} \mathrm{of} \mathrm{the} \mathrm{cloth} (\mathrm{per} \mathrm{metre}) = \mathrm{Rs} 31\frac{1}{2}\\ \end{array}$